Question

If $\int \frac{2Cosx-3Sinx}{3Sinx+2Cosx}dx=A\log |3sinx+$$2cosx|+BX+C$, then $A=\dots \dots \dots .,B=\dots \dots \dots \dots$,

• A. $\frac{12}{13},-\frac{5}{13}$
• B. $\frac{1}{13},-\frac{5}{13}$
• C. $12,-\frac{5}{13}$
• D. $\frac{12}{13},\frac{5}{13}$

Answer 1

Answer: (A)

$\frac{12}{13},-\frac{5}{13}$

$\int \frac{2cosx-3\sin x}{3sinx+2cosx}dx$
we will write the function in the numerator in the form
$N\left(x\right)=\lambda \left(D^{\prime }\right(x\left)\right)+\mu \left(D\right(x\left)\right)$
Where N(x) is the numerator function and D(x)is denominator function.
$2cosx-3sinx=\lambda (3cosx-2\sin x)+\mu (3sinx+2cosx)$
$\begin{array}{ccc}2=& 3\lambda +& 2\mu \\ -3=& -2\lambda +& 3\mu \end{array}\}\mu =-\frac{5}{13};\lambda =\frac{12}{13}$
$N\left(x\right)=\frac{12}{13}\left(D^{{}^{\prime }}\right(x\left)\right)-\frac{5}{13}D\left(x\right)$
$\int \frac{N\left(x\right)}{D\left(x\right)}dx=$ $\frac{12}{3}\int \frac{D^{\prime }\left(x\right)dx}{D\left(x\right)}-\frac{5}{13}\int \frac{D\left(x\right)}{D\left(x\right)}dx$
$=\frac{12}{13}log\left|\right(D\left(x\right)\left)\right|-\frac{5}{13}x+c$
$=\frac{12}{13}log(3sinx+2cosx)-\frac{5}{13}x+c$
$A=\frac{12}{13}$; $B=\frac{-5}{13}$